Chapter 3 · Matrices

Example 24: For symmetric $A, B$ ⁠: $AB$ symmetric iff $AB=BA$

ExamplesMiscellaneous examples · Chapter 3

The question

If $A$ and $B$ are symmetric matrices of the same order, then show that $AB$ is symmetric if and only if $A$ and $B$ commute, that is $AB = BA$ ⁠.

Exam answer

Given $A' = A$ and $B' = B$ ⁠. So

(AB)' = B'A' = BA \quad\ldots(\text{i})

Let $AB$ be symmetric, that is $(AB)' = AB$ ⁠. Then by (i), $BA = AB$ ⁠.

Conversely, let $AB = BA$ ⁠. Then by (i),

(AB)' = BA = AB

so $AB$ is symmetric.

Hence $AB$ is symmetric if and only if $AB = BA$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Transposing a product reverses it: $(AB)' = B'A'$ ⁠. For symmetric matrices $B' = B$ and $A' = A$ ⁠, so $(AB)' = BA$ ⁠. That single line says: the transpose of $AB$ is $BA$ . Now " $AB$ is symmetric" means $(AB)' = AB$ ⁠, which is the same as $BA = AB$ ⁠. Both directions of the "if and only if" come out of this one identity.

Given, to show, and the plan

Given: Square matrices $A$ and $B$ of the same order $n\times n$ ⁠, with $A' = A$ and $B' = B$ ⁠.
To show: Both directions: (1) if $AB$ is symmetric, then $AB = BA$ ⁠; (2) if $AB = BA$ ⁠, then $AB$ is symmetric.
Plan: Check that everything is defined; establish $(AB)' = BA$ ⁠; prove each direction from it; then look at examples showing both outcomes, and at what goes wrong when $A$ ⁠, $B$ are not symmetric.

Basics you need

1Symmetric matrix

A square matrix $S$ is symmetric if $S' = S$ ⁠, that is $s_{ji} = s_{ij}$ for all $i, j$ ⁠. Basics: mirror patterns

2Transpose of a product

$(AB)' = B'A'$ whenever $AB$ is defined: the order of the factors reverses. (Section 3.5.1, property (iv); a proof is given in Step 2.)

3"If and only if"

" $X$ if and only if $Y$ " is two statements: "if $Y$ then $X$ " and "if $X$ then $Y$ ". A complete proof proves both directions separately.

4Matrix products need not commute

For matrices, $AB$ and $BA$ are often different, even when both are defined. " $A$ and $B$ commute" means the special case $AB = BA$ ⁠. Basics: pair up, multiply, add

Solution

Check that $AB$ and $BA$ are defined

A symmetric matrix must be square (otherwise $A'$ would have a different order from $A$ and could not equal it). Since $A$ and $B$ have the same order, both are $n\times n$ for some $n$ ⁠. Then $AB$ is defined (columns of $A$ = $n$ = rows of $B$ ⁠) and is $n\times n$ ⁠; likewise $BA$ is defined and is $n\times n$ ⁠. So it makes sense to compare $AB$ with $BA$ ⁠, and to ask whether the square matrix $AB$ is symmetric.

" $AB = BA$ " can only hold between matrices of the same order, and "symmetric" only applies to square matrices. Checking orders first makes sure the statement we are proving is meaningful.

Definition 4, Section 3.6 (symmetric matrices are square); Section 3.4.5 (orders of a product). Lesson: symmetric and skew symmetric
The key identity: $(AB)' = BA$

By the reversal rule and then the symmetry of $A$ and $B$ ⁠,
$(AB)' = B'A' = BA.$
Why the reversal rule holds: the $(i,j)$ entry of $(AB)'$ is the $(j,i)$ entry of $AB$ ⁠, which is
$\textstyle\sum_{r=1}^{n} a_{jr}\,b_{ri}.$
The $(i,j)$ entry of $B'A'$ is (row $i$ of $B'$ ⁠) times (column $j$ of $A'$ ⁠), and row $i$ of $B'$ is column $i$ of $B$ ⁠, column $j$ of $A'$ is row $j$ of $A$ ⁠. So it is
$\textstyle\sum_{r=1}^{n} b_{ri}\,a_{jr},$
the same numbers multiplied in the other order. Hence $(AB)' = B'A'$ for every $i, j$ ⁠.

This answers the textbook's "(Why?)". The first equality is the reversal rule; the second replaces $B'$ by $B$ and $A'$ by $A$ ⁠, which is allowed because both matrices are symmetric (Given). Everything else in the proof rests on this line.

Property (iv), Section 3.5.1; see also Example 21. Lesson: transpose
Direction 1: if $AB$ is symmetric, then $AB = BA$

Assume $AB$ is symmetric, so $(AB)' = AB$ ⁠. By Step 2, $(AB)' = BA$ ⁠. Both are equal to $(AB)'$ ⁠, so
$BA = (AB)' = AB.$
Thus $AB = BA$ : $A$ and $B$ commute.

We start from the assumption of this direction and use only it, the givens, and the identity of Step 2. Two things equal to the same thing are equal to each other.
Direction 2: if $AB = BA$ ⁠, then $AB$ is symmetric

Assume $AB = BA$ ⁠. By Step 2 and then this assumption,
$(AB)' = BA = AB.$
So $(AB)' = AB$ ⁠, which says exactly that $AB$ is symmetric.

This is the "conversely" part of the textbook solution. Again we start from this direction's own assumption. The two directions together prove the "if and only if".

Definition 4, Section 3.6: $S$ is symmetric when $S' = S$ ⁠.
See both cases in examples

A pair that does not commute. Take the symmetric $2\times2$ matrices
$A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$
Entry by entry,
$\begin{aligned} (AB)_{11} &= 1\cdot0 + 0\cdot1 = 0,\\ (AB)_{12} &= 1\cdot1 + 0\cdot0 = 1,\\ (AB)_{21} &= 0\cdot0 + 2\cdot1 = 2,\\ (AB)_{22} &= 0\cdot1 + 2\cdot0 = 0,\\ (BA)_{11} &= 0\cdot1 + 1\cdot0 = 0,\\ (BA)_{12} &= 0\cdot0 + 1\cdot2 = 2,\\ (BA)_{21} &= 1\cdot1 + 0\cdot0 = 1,\\ (BA)_{22} &= 1\cdot0 + 0\cdot2 = 0. \end{aligned}$ $AB = \begin{bmatrix} 0 & 1 \\ 2 & 0 \end{bmatrix}, \qquad BA = \begin{bmatrix} 0 & 2 \\ 1 & 0 \end{bmatrix}.$
Here $AB \ne BA$ ⁠, and $AB$ is not symmetric ( $1 \ne 2$ across the diagonal). Notice that $BA$ is exactly $(AB)'$ ⁠, as Step 2 says.

A pair that commutes. Now take
$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, \qquad B = \begin{bmatrix} 2 & 2 \\ 2 & 4 \end{bmatrix},$
both symmetric ( $B$ is $A + I$ ⁠). Entry by entry,
$\begin{aligned} (AB)_{11} &= 1\cdot2 + 2\cdot2 = 6,\\ (AB)_{12} &= 1\cdot2 + 2\cdot4 = 10,\\ (AB)_{21} &= 2\cdot2 + 3\cdot2 = 10,\\ (AB)_{22} &= 2\cdot2 + 3\cdot4 = 16,\\ (BA)_{11} &= 2\cdot1 + 2\cdot2 = 6,\\ (BA)_{12} &= 2\cdot2 + 2\cdot3 = 10,\\ (BA)_{21} &= 2\cdot1 + 4\cdot2 = 10,\\ (BA)_{22} &= 2\cdot2 + 4\cdot3 = 16. \end{aligned}$ $AB = BA = \begin{bmatrix} 6 & 10 \\ 10 & 16 \end{bmatrix},$
which is symmetric, as the theorem predicts.

The examples are not part of the proof (the proof is Steps 3 and 4), but they show that both situations really occur, so neither condition is automatic for symmetric matrices.
Why the word "symmetric" in the hypothesis matters

Without it, Direction 1 can fail. Take the (non-symmetric) matrices
$A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}.$
Entry by entry,
$\begin{aligned} (AB)_{11} &= 0\cdot0 + 1\cdot1 = 1,\\ (AB)_{12} &= 0\cdot0 + 1\cdot0 = 0,\\ (AB)_{21} &= 0\cdot0 + 0\cdot1 = 0,\\ (AB)_{22} &= 0\cdot0 + 0\cdot0 = 0,\\ (BA)_{11} &= 0\cdot0 + 0\cdot0 = 0,\\ (BA)_{12} &= 0\cdot1 + 0\cdot0 = 0,\\ (BA)_{21} &= 1\cdot0 + 0\cdot0 = 0,\\ (BA)_{22} &= 1\cdot1 + 0\cdot0 = 1. \end{aligned}$ $AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \qquad BA = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.$
$AB$ is symmetric, yet $AB \ne BA$ ⁠. The proof breaks at Step 2: here $B'A' = BA$ is false, because $A' \ne A$ and $B' \ne B$ ⁠.

A counterexample is the right tool for showing that a statement is false in general. It also shows exactly which given the proof depends on.

Result

For symmetric $A$ ⁠, $B$ of the same order, $(AB)' = B'A' = BA$ ⁠. Hence $AB$ is symmetric $\iff (AB)' = AB \iff BA = AB$ ⁠. Both directions are proved: $AB$ is symmetric if and only if $A$ and $B$ commute.

Check it with numbers

A computer test over 2000 random pairs of symmetric $3\times3$ integer matrices (and many specially built commuting pairs, such as $B = A^2 - 2A + 3I$ ⁠) found that " $AB$ is symmetric" and " $AB = BA$ " were either both true or both false in every case, and that $(AB)' = BA$ every time.

Common mistakes

Proving only one direction. "If and only if" needs both. The textbook marks the second with "Conversely".
Writing $(AB)' = A'B'$ ⁠. The order reverses: $(AB)' = B'A'$ ⁠.
Thinking symmetric matrices always commute. The first pair in Step 5 is symmetric but $AB \ne BA$ ⁠.
Using the hypotheses in the wrong direction. In Direction 1 you may assume $(AB)' = AB$ ⁠; in Direction 2 you may assume $AB = BA$ ⁠. Do not use the conclusion of a direction inside its own proof.

What to take away

For symmetric $A$ ⁠, $B$ ⁠: the transpose of $AB$ is $BA$ ⁠.
So $AB$ is symmetric exactly when $AB = BA$ ⁠.
Prove "if and only if" as two separate implications, each starting from its own assumption.

Every fact used, and where it comes from

Fact	Used in	Source
Symmetric: $S' = S$ (so $S$ is square)	Steps 1 to 4	Definition 4, Section 3.6 (Lesson: symmetric and skew symmetric)
Orders of a product; entry rule	Steps 1, 2, 5, 6	Section 3.4.5 (Lesson: matrix multiplication)
$(AB)' = B'A'$	Step 2	Section 3.5.1, property (iv) (Lesson: transpose)
$AB \ne BA$ in general	Steps 5, 6	Section 3.4.5, non-commutativity (Lesson: properties)
Proof of both directions	Steps 3, 4	Miscellaneous Examples, Example 24

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Example 24: For symmetric A,BA, BA,B⁠: ABABAB symmetric iff AB=BAAB=BAAB=BA

Basics you need

Solution

Check that ABABAB and BABABA are defined

The key identity: (AB)′=BA(AB)' = BA(AB)′=BA

Direction 1: if ABABAB is symmetric, then AB=BAAB = BAAB=BA

Direction 2: if AB=BAAB = BAAB=BA⁠, then ABABAB is symmetric

See both cases in examples

Why the word "symmetric" in the hypothesis matters

Every fact used, and where it comes from

Example 24: For symmetric $A, B$ ⁠: $AB$ symmetric iff $AB=BA$

Check that $AB$ and $BA$ are defined

The key identity: $(AB)' = BA$

Direction 1: if $AB$ is symmetric, then $AB = BA$

Direction 2: if $AB = BA$ ⁠, then $AB$ is symmetric